The problem goes as follows: "Let $U$ be an open subset of $\mathbb{R}^N$ and $f: U \to \mathbb{R}^k$ a $C^{\infty}$ map. If zero is a regular value of $f$, the set $X = f^{-1}(0)$ is a manifold of dimension $n = N-k$. Show that this manifold has a natural smooth orientation."
The first hint of the problem then states "Let $f = (f_1, \cdots, f_k)$ and let $df_1 \wedge \cdots \wedge df_k = \sum_I f_Idx_I$ summed over the multi-indices which are strictly increasing. Show that for every $p \in X$, $f_I(p) \neq 0$ for some multi-index $I, (i_1, \cdots, i_k), 1 \leq i_1 < \cdots < i_k \leq N$."
I am very confused by multiple parts of this hint. Shouldn't $df_1 \wedge \cdots \wedge df_k = \mbox{det}\left(\frac{\partial f_j}{\partial x_i}\right)dx_1 \wedge \cdots \wedge dx_n$? What is $f_I$ in this case? And how can the index go up through $N$ if there all only $k$ component functions? I am assuming its due to the partials but want to be sure. The book my class is using has a lot of typos, so I am assuming that's what this boils down to. Want to clarify though. Thank you!
First of all $\displaystyle df_{j}=\sum_{i=1}^{N}\frac{\partial f_{j}}{\partial x_{i}}dx_{i}\,,j=1,2,...k$
So $\displaystyle df_{1}\wedge df_{2}...\wedge df_{k}=\bigg(\sum_{i=1}^{N}\frac{\partial f_{1}}{\partial x_{i}}dx_{i}\bigg)\wedge\bigg(\sum_{i=1}^{N}\frac{\partial f_{2}}{\partial x_{i}}dx_{i}\bigg)\wedge\dots\wedge\bigg(\sum_{i=1}^{N}\frac{\partial f_{k}}{\partial x_{i}}dx_{i}\bigg)$
So in particular it would have terms like $(\frac{\partial f_{1}}{\partial x_{i_{1}}}dx_{i_{1}})\wedge(\frac{\partial f_{2}}{\partial x_{i_{2}}}dx_{i_{2}})\wedge\dots\wedge(\frac{\partial f_{k}}{\partial x_{i_{k}}}dx_{i_{k}})$. So when you write them in their unique representation you WILL get the form $\displaystyle\sum_{I}^{\text{increasing}}f_{I}\cdot dx_{I}$.
Now if you do prove that $ df_{1}\wedge df_{2}...\wedge df_{k}$ is a nowhere vanishing on the submanifold then it get's its natural orientation from it.
Here is a complete solution to the problem:-
By the Regular Level Set Theorem you have that $f^{-1}(0)$ is a submanifold of the required codimension.
Now as $0$ is a regular value, we have for every point in the submanifold , the matrix $\bigg[\frac{\partial f_{i}}{\partial x_{j}}\bigg]_{k\times N}$ has full rank $=k$. And hence in particular $\{df_{i}\}_{i=1}^{k}$ is a linearly independent set.
Now it is easy to prove that if $\omega_{1},...,\omega_{k}$ are $1$-forms on a finite dimensional vector space $V$ (with dimension not necessarily $k$) , then $\omega_{1}\wedge\dots\wedge\omega_{k}=0$ if and only if $\omega_{1},...,\omega_{k}$ are linearly dependent.
One direction is obvious by just writing $\displaystyle\omega_{k}=\sum_{i=1}^{k-1}c_{i}\omega_{i}$ and then expanding the wedge product and using the fact that the $\omega_{i}\wedge\omega_{i}=0$ .
And conversely if they are linearly independent then you extend them to a basis for the dual space $V^*$ and then use the basis to get the corresponding double dual basis for $V^{**}$ and use the canonical identification of the vector space $V$ with it's double dual to get $k$ linearly independent vectors $E_{1},...,E_{k}$ such that $\omega_{i}(E_{j})=\delta_{ij}$ where $\delta_{ij}$ is the Kronecker Delta.
Now it is again an easy excercise to prove that $\omega_{1}\wedge\dots\wedge\omega_{k}(v_{1},...,v_{k})=\det(\omega_{i}(v_{j}))$ .
Thus we have $\omega_{1}\wedge\dots\wedge\omega_{k}(E_{1},...,E_{k})=\det(I_{k\times k})=1\neq 0$ and thus $\omega_{1}\wedge\dots\omega_{k}\neq 0$.
Hence using these above results you can conclude as required.